Question

Question 2: The Rapid Test is used to determine whether someone has HIV (the virus that causes
AIDS). The falsepositive and false-negative rates are .028 and .091, respectively. A physician has
just received the Rapid Test report that his patient tested positive. Before receiving the result, the
physician assigned his patient to the low-risk group (defined on the basis of several variables) with
only a 0.6% probability of having HIV. What is the probability that the patient actually has HIV?

Answer 1

Answer:

0.1626 = 16.26% probability that the patient actually has HIV

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Positive test

Event B: Has HIV.

Probability of a positive test:

0.028 of 100 - 0.6 = 99.4% = 0.994(false-positive).

1 - 0.091 = 0.901 out of 0.6% = 0.006(positive). So

$P(A) = 0.028*0.994 + 0.901*0.006 = 0.033238$

Probability of a positive test and having HIV:

0.901 out of 0.006. So

$P(A \cap B) = 0.901*0.006 = 0.005406$

What is the probability that the patient actually has HIV?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.005406}{0.033238} = 0.1626$

0.1626 = 16.26% probability that the patient actually has HIV